For example object with v=tanh(0.1213)t/0.01236. m=69kg.
I thought I had to use F=mv/t but that just gives the net force of the object while falling.
I don't know how to find the force impacting on the ground.
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3$\begingroup$ you need more info, like how long it takes to decelerate due to the ground $\endgroup$– Pato GalmariniCommented Jul 8 at 5:04
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$\begingroup$ what formula for the impact force should I use $\endgroup$– Yifan YINCommented Jul 8 at 5:06
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$\begingroup$ F=ma................................... $\endgroup$– Pato GalmariniCommented Jul 8 at 5:10
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$\begingroup$ that doesnt work, and that is for constant acceleration $\endgroup$– Yifan YINCommented Jul 8 at 5:11
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3$\begingroup$ Well, you need to know how long the object is in contact with the ground before it comes to complete rest (assuming no rebound). $\endgroup$– joseph hCommented Jul 8 at 5:23
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2 Answers
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Short answer: The given information is not enough.
Long answer: You correctly noted that the force of the impact is related to the change in momentum. We take off from there. If we assume the duration of impact to be $\Delta t$, then we have: $$F \Delta t = \Delta mv = m \Delta v$$ (the mass of the object does not change in impact)
From the above equation we can see that in addition to $m$, we need two other pieces of information in order to calculate impact force: $\Delta t$ (duration of impact) and $\Delta v$ (change of velocity in impact). The latter may seem a bit confusing, so I expand on it a bit more here.
Imagine that the falling object is a pile of mud. It reaches ground with velocity $v$ and immediately after impact its velocity is $0$. This type of impact is called inelastic (or sometimes called plastic). In this case, $\Delta v = 0 - v = -v$.
Now imagine another object of the same mass and the same velocity $v$ at the beginning of impact. This time, however, suppose that the object bounces back from the ground with velocity $-v$, somewhat like a pingpong ball. This type of impact is called elastic. In this case, $\Delta v = -v - v = -2v$.
Compare the above examples of inelastic and elastic collisions. Note that even if $m$, $v$ and $\Delta t$ are the same in the two examples, their forces of impact differ, because the two cases involve different changes in velocity, and therefore different changes in momentum.
As a final note, collisions in reality are not necessarily perfectly inelastic or perfectly elastic; they may be somewhere in between. In such a case, an additional piece of information that is needed in order to calculate the force of impact is a coefficient quantifying the elasticity of the impact.
Edit: The term inelastic collision is used more frequently than plastic collision. Following Cameron Williams' comment I replaced the word plastic with inelastic.
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1$\begingroup$ I have heard the term "completely inelastic" rather than "plastic" in this situation. $\endgroup$ Commented Jul 8 at 6:08
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$\begingroup$ @CameronWilliams Thank you. I have heard them both (see for example Britannica entry for 'inelastic collision') but the term 'inelastic' is used more often. So I edited the text following your comment. $\endgroup$– SaeedCommented Jul 8 at 12:10
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You need to know either impact time, as per $$ F = -\frac {p_f}{t_i} $$
Or you need to know impact distance as in, for example meteorite+crater case : $$ F=\frac {K_f}{d_i} $$
when equating meteorite kinetic energy to work done by impact force,- we can extrapolate force.
Without any impact parameters, it's not possible to calculate force exerted on the ground.
answered Jul 8 at 6:44